Was the Demon Core Hot?
Adapted and expanded from my answer on Physics Stack Exchange.
In 1945, Los Alamos cast a 6.2-kilogram sphere of plutonium to be the core of a third atomic bomb. Japan surrendered before it was needed, and the sphere stayed at the lab as a test object for criticality experiments. Over the next year it killed two physicists: Harry Daghlian died in 1945 after fumbling a tungsten-carbide brick onto it; Louis Slotin died in 1946 when a screwdriver slipped and let two beryllium hemispheres close around it. Both accidents drove the assembly briefly supercritical and flooded the room with neutrons and gamma rays. Thereafter, it began to be known as the demon core.
Set aside the criticality scenarios for a moment. A plutonium pit is radioactive, which means that even sitting inert on a bench it is continuously converting nuclear binding energy into heat. This raises an interesting question – just resting there, would the demon core have felt hot to the touch?
Here is how we can find out.
How much heat did it produce?
Plutonium-239 dissipates roughly \(1.9~\mathrm{W\,kg^{-1}}\) through its alpha decay.Weapons-grade plutonium is not pure Pu-239. It carries a few percent Pu-240 (\(\sim7~\mathrm{W\,kg^{-1}}\)) and traces of Pu-238 (\(\sim560~\mathrm{W\,kg^{-1}}\)), both of which run hotter. So this is a mild under-estimate – the real pit would have generated slightly more power than we calculate here. For our 6.2-kg sphere, that gives a steady thermal output of
$$P \approx 1.9~\mathrm{W\,kg^{-1}} \times 6.2~\mathrm{kg} \approx 12~\mathrm{W}.$$
About the heat of a dim incandescent night-light, generated continuously. There is nowhere for this heat to escape, other than across the surface of the sphere.
How big is the surface?
Plutonium is dense, around \(19.8~\mathrm{g\,cm^{-3}}\), so 6.2 kg occupies only \(V \approx 313~\mathrm{cm^3}\).The real pit was gallium-stabilized delta-phase plutonium, which is less dense (\(\sim15.9~\mathrm{g\,cm^{-3}}\)) and so slightly larger – about a 9-cm sphere. This only shifts things a few degrees, though, and we are only interested in the coarse behavior (is it “hot”), so I’m not going to complicate things. A sphere of that volume has radius
$$r = \left(\frac{3V}{4\pi}\right)^{1/3} \approx 4.2~\mathrm{cm},$$
and therefore a surface area
$$A = 4\pi r^2 \approx 223~\mathrm{cm^2} = 0.0223~\mathrm{m^2}.$$
This means the sphere was just a little bigger than a baseball.
Equilibrium by radiation alone
In a vacuum, the only way out for the heat is thermal radiation. At equilibrium the radiated power must balance the decay heat (if it didn’t, the temperature would be changing and so it wouldn’t be equilibrium), so by the Stefan-Boltzmann law we have
$$\varepsilon\sigma A\left(T^4 - T_0^4\right) = P,$$
where \(\sigma = 5.67\times10^{-8}~\mathrm{W\,m^{-2}\,K^{-4}}\) is the Stefan-Boltzmann constant, \(T_0 = 293~\mathrm{K}\) is the ambient temperature (20 °C), \(\varepsilon = 1\) is the emissivityThe emissivity is basically a measure of how efficiently an object emits thermal radiation. Here we take it to be 1 as an upper bound. Lowering it results in a higher final temperature as the object can’t reject heat via radiation as effectively., and \(P\) is the generated heat. Solving for the surface temperature, we find
$$\begin{aligned}T &= \left(T_0^4 + \frac{P}{\varepsilon\sigma A}\right)^{1/4} \\ &\approx \left(7.4\times10^{9} + 9.5\times10^{9}\right)^{1/4}~\mathrm{K} \\ &\approx 360~\mathrm{K}\end{aligned}$$
or about 87 °C. This is definitely hot enough to be uncomfortable to hold, on the order of a fresh cup of coffee.
But it wasn’t in a vacuum
The accidents happened in air, so the core also loses heat by convection. A crude modelThis model assumes that the power dissipated by convection is linear in the temperature difference between the object and the environment, and linear in the exposed surface area. adds a term \(P_\text{conv} = hA(T - T_0)\) with a natural-convection coefficient of order \(h \approx 10~\mathrm{W\,m^{-2}\,K^{-1}}\). Balancing radiation and convection against the same 12 watts,
$$\varepsilon\sigma A\left(T^4 - T_0^4\right) + hA\left(T - T_0\right) = P,$$
drops the equilibrium temperature to roughly 50 °C – about as warm as a hot bath, or that cup of coffee ten minutes after you poured it.
So, was it hot?
Well, I guess that depends on perspective. Somewhere in the neighborhood of body temperature to a hot drink, depending on how freely it could shed heat to the air around it (or objects it was resting against). That matches with accounts from the people who actually handled plutonium pits, who described them as faintly, eerily warm to the touch – one famous description likened the sensation to holding a live rabbit.
If you are going to be handling plutonium pits, you can at the very least rest assured that you shouldn’t be (thermally) burned by them.